Question

RESOLVED:

A quantity of 25.0 mL of a solution containing both Fe2+ and Fe3+ ions is titrated with 23.0 mL of 0.0200 M KMnO4 (in dilute sulfuric acid). As a result, all of the Fe2+ ions are oxidized to Fe3+ ions. Next, the solution is treated with Zn metal to convert all of the Fe3+ ions to Fe2+ ions. Finally, the solution containing only the Fe2+ ions requires 40.0 mL of the same KMnO4 solution for oxidation to Fe3+. Calculate the molar concentrations of Fe2+ and Fe3+ in the original solution. The net ionic equation is MnO4- + 5Fe2+ + 8H+ >>> Mn2+ + 5Fe3+ + 4H2O.

Answer 1

Let's start by analyzing the given information and using the net ionic equation to determine the molar concentrations of Fe2+ and Fe3+ in the original solution.

From the net ionic equation:

MnO4- + 5Fe2+ + 8H+ → Mn2+ + 5Fe3+ + 4H2O

We can see that 1 mole of MnO4- reacts with 5 moles of Fe2+ to produce 5 moles of Fe3+.

Given:

Volume of KMnO4 solution used = 23.0 mL = 0.023 L

Concentration of KMnO4 solution = 0.0200 M

Using the equation:

C1V1 = C2V2

Where:

C1 = concentration of KMnO4 solution

V1 = volume of KMnO4 solution used

C2 = concentration of Fe2+ in the original solution

V2 = volume of Fe2+ solution (unknown)

We can calculate the moles of KMnO4 used:

moles of KMnO4 = C1 * V1

moles of KMnO4 = 0.0200 M * 0.023 L

Since 1 mole of MnO4- reacts with 5 moles of Fe2+, the moles of Fe2+ in the original solution are equal to the moles of KMnO4 used.

moles of Fe2+ = moles of KMnO4 = 0.0200 M * 0.023 L

Next, we need to determine the moles of Fe3+ produced. From the net ionic equation, we know that 1 mole of MnO4- reacts with 5 moles of Fe2+ to produce 5 moles of Fe3+.

moles of Fe3+ = 5 * moles of Fe2+ = 5 * (0.0200 M * 0.023 L)

Now, let's consider the second part of the problem, where the solution containing only Fe2+ requires 40.0 mL of the same KMnO4 solution for oxidation to Fe3+.

Using the same equation:

C1V1 = C2V2

Where:

C1 = concentration of KMnO4 solution

V1 = volume of KMnO4 solution used

C2 = concentration of Fe2+ in the solution

V2 = volume of Fe2+ solution (unknown)

We can calculate the moles of KMnO4 used in the second part:

moles of KMnO4 = C1 * V1

moles of KMnO4 = 0.0200 M * 0.040 L

Since 1 mole of MnO4- reacts with 1 mole of Fe2+, the moles of Fe2+ in the solution are equal to the moles of KMnO4 used.

moles of Fe2+ = moles of KMnO4 = 0.0200 M * 0.040 L

Now, we can calculate the moles of Fe3+ produced in the second part. From the net ionic equation, we know that 1 mole of MnO4- reacts with 5 moles of Fe2+ to produce 5 moles of Fe3+.

moles of Fe3+ = 5 * moles of Fe2+ = 5 * (0.0200 M * 0.040 L)

Finally, we can calculate the molar concentrations of Fe2+ and Fe3+ in the original solution.

Concentration of Fe2+ = moles of Fe2+ / volume of Fe2+ solution

Concentration of Fe3+ = moles of Fe3+ / volume of Fe2+ solution

Since the volumes of Fe2+ solution in both parts are the same, we can cancel out the volume term.

Concentration of Fe2+ = moles of Fe2+ / volume of Fe2+ solution = 0.0200 M * 0.023 L / V2

Concentration of Fe3+ = moles of Fe3+ / volume of Fe2+ solution = 5 * (0.0200 M * 0.023 L) / V2

Therefore, the molar concentrations of Fe2+ and Fe3+ in the original solution are 0.0200 M * 0.023 L / V2 and 5 * (0.0200 M * 0.023 L) / V2, respectively.